3.237 \(\int \frac{A+B x^3}{x^3 (a+b x^3)^{3/2}} \, dx\)

Optimal. Leaf size=272 \[ -\frac{\sqrt{2+\sqrt{3}} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} (7 A b-4 a B) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right ),-7-4 \sqrt{3}\right )}{6 \sqrt [4]{3} a^2 \sqrt [3]{b} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}-\frac{x (7 A b-4 a B)}{6 a^2 \sqrt{a+b x^3}}-\frac{A}{2 a x^2 \sqrt{a+b x^3}} \]

[Out]

-A/(2*a*x^2*Sqrt[a + b*x^3]) - ((7*A*b - 4*a*B)*x)/(6*a^2*Sqrt[a + b*x^3]) - (Sqrt[2 + Sqrt[3]]*(7*A*b - 4*a*B
)*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2
]*EllipticF[ArcSin[((1 - Sqrt[3])*a^(1/3) + b^(1/3)*x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)], -7 - 4*Sqrt[3]])/
(6*3^(1/4)*a^2*b^(1/3)*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3)*x))/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*Sqrt[a + b*
x^3])

________________________________________________________________________________________

Rubi [A]  time = 0.100817, antiderivative size = 272, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {453, 199, 218} \[ -\frac{\sqrt{2+\sqrt{3}} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} (7 A b-4 a B) F\left (\sin ^{-1}\left (\frac{\sqrt [3]{b} x+\left (1-\sqrt{3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt{3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt{3}\right )}{6 \sqrt [4]{3} a^2 \sqrt [3]{b} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}-\frac{x (7 A b-4 a B)}{6 a^2 \sqrt{a+b x^3}}-\frac{A}{2 a x^2 \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/(x^3*(a + b*x^3)^(3/2)),x]

[Out]

-A/(2*a*x^2*Sqrt[a + b*x^3]) - ((7*A*b - 4*a*B)*x)/(6*a^2*Sqrt[a + b*x^3]) - (Sqrt[2 + Sqrt[3]]*(7*A*b - 4*a*B
)*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2
]*EllipticF[ArcSin[((1 - Sqrt[3])*a^(1/3) + b^(1/3)*x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)], -7 - 4*Sqrt[3]])/
(6*3^(1/4)*a^2*b^(1/3)*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3)*x))/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*Sqrt[a + b*
x^3])

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rubi steps

\begin{align*} \int \frac{A+B x^3}{x^3 \left (a+b x^3\right )^{3/2}} \, dx &=-\frac{A}{2 a x^2 \sqrt{a+b x^3}}-\frac{\left (\frac{7 A b}{2}-2 a B\right ) \int \frac{1}{\left (a+b x^3\right )^{3/2}} \, dx}{2 a}\\ &=-\frac{A}{2 a x^2 \sqrt{a+b x^3}}-\frac{(7 A b-4 a B) x}{6 a^2 \sqrt{a+b x^3}}-\frac{(7 A b-4 a B) \int \frac{1}{\sqrt{a+b x^3}} \, dx}{12 a^2}\\ &=-\frac{A}{2 a x^2 \sqrt{a+b x^3}}-\frac{(7 A b-4 a B) x}{6 a^2 \sqrt{a+b x^3}}-\frac{\sqrt{2+\sqrt{3}} (7 A b-4 a B) \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right )|-7-4 \sqrt{3}\right )}{6 \sqrt [4]{3} a^2 \sqrt [3]{b} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}\\ \end{align*}

Mathematica [C]  time = 0.0348464, size = 86, normalized size = 0.32 \[ \frac{x^3 \sqrt{\frac{b x^3}{a}+1} (4 a B-7 A b) \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{b x^3}{a}\right )-6 a A+8 a B x^3-14 A b x^3}{12 a^2 x^2 \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^3)/(x^3*(a + b*x^3)^(3/2)),x]

[Out]

(-6*a*A - 14*A*b*x^3 + 8*a*B*x^3 + (-7*A*b + 4*a*B)*x^3*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[1/3, 1/2, 4/3, -
((b*x^3)/a)])/(12*a^2*x^2*Sqrt[a + b*x^3])

________________________________________________________________________________________

Maple [B]  time = 0.021, size = 631, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)/x^3/(b*x^3+a)^(3/2),x)

[Out]

B*(2/3/a*x/((x^3+a/b)*b)^(1/2)-2/9*I/a*3^(1/2)/b*(-a*b^2)^(1/3)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a
*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)*((x-1/b*(-a*b^2)^(1/3))/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-
a*b^2)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/
2)/(b*x^3+a)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/
(-a*b^2)^(1/3))^(1/2),(I*3^(1/2)/b*(-a*b^2)^(1/3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2
)))+A*(-1/2/a^2*(b*x^3+a)^(1/2)/x^2-2/3*b/a^2*x/((x^3+a/b)*b)^(1/2)+7/18*I/a^2*3^(1/2)*(-a*b^2)^(1/3)*(I*(x+1/
2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)*((x-1/b*(-a*b^2)^(1/3))/(-3
/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2
)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*
I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),(I*3^(1/2)/b*(-a*b^2)^(1/3)/(-3/2/b*(-a*b^2)^(1/3)
+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac{3}{2}} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^3/(b*x^3+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x^3 + A)/((b*x^3 + a)^(3/2)*x^3), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B x^{3} + A\right )} \sqrt{b x^{3} + a}}{b^{2} x^{9} + 2 \, a b x^{6} + a^{2} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^3/(b*x^3+a)^(3/2),x, algorithm="fricas")

[Out]

integral((B*x^3 + A)*sqrt(b*x^3 + a)/(b^2*x^9 + 2*a*b*x^6 + a^2*x^3), x)

________________________________________________________________________________________

Sympy [A]  time = 39.287, size = 82, normalized size = 0.3 \begin{align*} \frac{A \Gamma \left (- \frac{2}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{2}{3}, \frac{3}{2} \\ \frac{1}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac{3}{2}} x^{2} \Gamma \left (\frac{1}{3}\right )} + \frac{B x \Gamma \left (\frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{3}{2} \\ \frac{4}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac{3}{2}} \Gamma \left (\frac{4}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**3+A)/x**3/(b*x**3+a)**(3/2),x)

[Out]

A*gamma(-2/3)*hyper((-2/3, 3/2), (1/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(3/2)*x**2*gamma(1/3)) + B*x*gamma(1/
3)*hyper((1/3, 3/2), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(3/2)*gamma(4/3))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac{3}{2}} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^3/(b*x^3+a)^(3/2),x, algorithm="giac")

[Out]

integrate((B*x^3 + A)/((b*x^3 + a)^(3/2)*x^3), x)